How do you find the volume of the solid generated by revolving the region bounded by the graphs #x=y^2, x=4#, about the line x=6?

1 Answer
Jim H
Mar 31, 2017

Please see below.

Explanation:

The graph of the region is shown below. A representative slice of thickness #dy# has been taken at height #y#. The radii of rotation are shown as dashed and dotted red lines.

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When revolved about the line #x=6# the resulting slice has volume

#(piR^2-pir^2) * "thickness"# where #R# is the greater radius and #r# the lesser.

In this problem

#R = 6-y^2# and
#r = 6-4=2# and
#"thickness" = dy#.

#y# varies from #-2# to #2#. (Or go from #0# to #2# and double the result.)

So the volume of the solid of revolution is

#V = pi int_-2^2 ((6-y^2)^2 - (2)^2) dy = pi int_-2^2 (32-12y^2+y^4) dy#

# = (384pi)/5#

or about #241.3#.

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